Question

$\underset{x\to -1}{lim}\frac{(1+x^{})(1-x^2)(1+x^3)(1-x^4)\cdots (1-x^{4n})}{\left[\right(1+x^{}\left)\right(1-x^2\left)\right(1+x^3\left)\right(1-x^4)\cdots (1-x^{2n}){]}^2}$ is equal to

• A. ${}^{4n}{C}_{2n}$
• B. ${}^{2n}{C}_n$
• C. $2\cdot {}^{4n}{C}_{2n}$
• D. $2\cdot {}^{2n}{C}_n$

Answer 1

The correct option is A ${}^{4n}{C}_{2n}$

$\underset{x\to -1}{lim}\frac{(1+x^{})(1-x^2)(1+x^3)(1-x^4)\cdots (1-x^{4n})}{\left[\right(1+x^{}\left)\right(1-x^2\left)\right(1+x^3\left)\right(1-x^4)\cdots (1-x^{2n}){]}^2}$

= $\underset{x\to -1}{lim}\frac{(1+x^{2n+1})}{1+x^{}}$x$\frac{(1-x^{2n+2})}{1-x^2}$x$\cdots$$\frac{(1-x^{4n})}{1-x^{2n}}$

= $\underset{x\to -1}{lim}\frac{(x^{2n+1}-(-1{)}^{2n+1})}{x^{}-(-1)}$x$\frac{(x^{2n+2}-(-1{)}^{2n+2})}{x^2-(-1{)}^2}$x$\cdots$$\frac{(x^{4n}-(-1{)}^{4n})}{x^{2n}-(-1{)}^{2n}}$
=$\frac{2n+1}{1}.\frac{2n+2}{2}.\frac{2n+3}{3}.\frac{2n+4}{4}....\frac{4n}{2n}={}^{4n}{C}_{2n}$